Question

DIFF EQ:
Is y = sin t a solution to (dy/dt)^2= 1 − y^2? Justify.

Answer 1

ill help

Answer 2

Find the second derivative of sin(t).

Answer 3

The reason I am asking you this, so that you can plug in the second derivative of sin(t) instead of (dy/dt)^2, and to plug in sin(t) instead of y.

Answer 4

oh

Answer 5

okay hold on @jigglypuff314

Answer 6

im srry

Answer 7

Just to clarify in case you didn't know (dy/dt)^2 is denoting the second derivative of the function y, with respect to t.

Answer 8

sorry, but what does omm stand for?
(not very good with English slang)

Answer 9

oh it mean didntt mean to say that

Answer 10

For example I have the following differential equation,
\(\color{#000000 }{ \displaystyle y''+2y'+y=2x+4 }\)

And I want to verify whether or not \(y=2x\) is a solution to the differential equation, than this is what I would do.

I know that,
\(\color{#000000 }{ \displaystyle y=2x }\)
\(\color{#000000 }{ \displaystyle y'=2 }\)
\(\color{#000000 }{ \displaystyle y''=0 }\)

And I will plug this into my differential equation,
\(\color{#000000 }{ \displaystyle (0)+2(2)+(2x)=2x+4 }\)
\(\color{#000000 }{ \displaystyle 4+2x=2x+4 }\)

So therefore, i have verified that \(\color{#000000 }{ \displaystyle y=2x }\)
is a solution to \(\color{#000000 }{ \displaystyle y''+2y'+y=2x+4 }\).

Answer 11

Thanks guys you were a big help when tried to solve for the equation my solution would always end up being y=-sin(t). I see what I did wrong now.

Answer 12

\(\color{#000000 }{ \displaystyle y''=1-y^2 }\)

If your solution is
\(\color{#000000 }{ \displaystyle y=a\sin(t) }\)

then,
\(\color{#000000 }{ \displaystyle y=a\sin(t) }\)
\(\color{#000000 }{ \displaystyle y'=a\cos(t) }\)
\(\color{#000000 }{ \displaystyle y''=-a\sin(t) }\)

and we plug that in,
\(\color{#000000 }{ \displaystyle y''=1-y^2\quad \Longrightarrow \quad -a\sin(t) =1-(a\sin(t))^2 }\)
and when you simplify you get,
\(\color{#000000 }{ \displaystyle -a\sin(t) =1-a^2\sin^2(t) }\)

\(\color{#000000 }{ \displaystyle a^2\sin^2(t) -a\sin(t)-1 =0}\)

and that is clearly not the solution

Answer 13

I know the solution is y = -sin(t), but how can I show this using the first order separable ODE method. Using the N(y)dy = M(x)dx , I know that N(y) = 1/(sqrt(-y^2+1)), but why?

Answer 14

Where'd the square root come from?

Answer 15

I know that I could show that y= -sin(t) by using the method you described but how could I do it with the First order separable method? Big thanks

Answer 16

If you differential equation, actually, is
\(\color{#000000 }{ \displaystyle (y')^2=1-y^2 }\)
(although I haven't ever seen derivatives
raised to a power in differential equations)

Then,
\(\color{#000000 }{ \displaystyle y'=\pm\sqrt{1-y^2} }\)

\(\color{#000000 }{ \displaystyle \pm\frac{1}{\sqrt{1-y^2}}\frac{dy}{dt}=1 }\)

\(\color{#000000 }{ \displaystyle \pm\color{red}{\int}\frac{1}{\sqrt{1-y^2}}\frac{dy}{dt}\color{red}{dt}=\color{red}{\int}\color{red}{dt} }\)

\(\color{#000000 }{ \displaystyle \pm\int\frac{1}{\sqrt{1-y^2}}dy=\int~dt }\)

\(\color{#000000 }{ \displaystyle y=\sin \theta }\)
\(\color{#000000 }{ \displaystyle dy=\sin \theta~ d\theta }\)

\(\color{#000000 }{ \displaystyle\pm \int\frac{1}{\cos \theta }dy=t+c }\)

and even in that case \(y(t)=\sin t\) is not a solution!

Answer 17

I have shown why \(\rm y(t)=a\sin(t) \) is not a solution for all "a"......

Answer 18

Could you look at this website...
https://www.symbolab.com/solver/ordinary-differential-equation-calculator/%5Cleft(%5Cfrac%7Bdy%7D%7Bdt%7D%5Cright)%5E%7B2%7D%3D1-y%5E%7B2%7D/?origin=button